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The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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26 Mar 2015, 03:36
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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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26 Mar 2015, 05:01
Bunuel wrote: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?
A. 102 B. 120 C. 132 D. 144 E. 156
Kudos for a correct solution. B and C should have same Y coordinate > \(4a5 = 2a+6\) > a=\(\frac{11}{2}\) area of triangle = \(\frac{1}{2} * (4a5) * (2a+1)\) . on substituting a=\(\frac{11}{2}\) we get area = 102 . answer A
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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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26 Mar 2015, 06:14
Bunuel wrote: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?
A. 102 B. 120 C. 132 D. 144 E. 156
Kudos for a correct solution. 1/2bh=1/2(2a+1)(2a+6) Now 4a5=2a+6 2a=11 Therefore, A(0,0); B(0,17); C(12,17) 1/2*17*12=102 Answer: A



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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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30 Mar 2015, 02:37



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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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12 Jan 2018, 11:22
AmoyV wrote: Bunuel wrote: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?
Now 4a5=2a+6
Sorry, but I did not get to your calculations  why it is so? What I missed?



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The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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19 Jan 2018, 01:13
Aksena wrote: AmoyV wrote: Bunuel wrote: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?
Now 4a5=2a+6
Sorry, but I did not get to your calculations  why it is so? What I missed? Im guessing you have problems understanding why 4a5 = 2a+6 Now, the reason they got to that equation was due to the given information that angle ABC = 90 > This implies that triangle ABC was a right angle > Now A was given at the origin (0,0), while B was given to be somewhere along the Yaxis as its Xcoordinates = 0 > Hence, as ABC was a right triangle, the point C will have the same Ycoordinates as B in order to form the right triangle (i.e. to fulfill angle A BC = 90) > B(0, 4a5) and C(2a+1, 2a+6) > Therefore 4a5 = 2a+6, solve for a and you can find the area of the triangle with the basic area formula for right triangles: base x height / 2



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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle
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21 Jan 2018, 18:33
Bunuel wrote: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?
A. 102 B. 120 C. 132 D. 144 E. 156 Since we have a right triangle with angle ABC = 90, we see that vertices A and B share the same xcoordinate, and vertices B and C share the same ycoordinate. Equating the ycoordinates of vertices B and C, we have: 4a  5 = 2a + 6 2a = 11 a = 5.5 Substituting 5.5 for a, we see that the ordered pair for vertex B is (0, 4(5.5)  5), or (0, 17). Thus, leg AB is the vertical distance from (0,0) to (0,17), or 17. Similarly, we see that the ordered pair for vertex C is (12, 17) and leg BC is is the horizontal distance from (0,17) to (12, 17) = 12. Thus, the area of the triangle is: area = base x height x 1/2 area = 17 x 12 x 1/2 = 17 x 6 = 102 Answer: A
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Re: The points A(0, 0), B(0, 4a  5) and C(2a + 1, 2a + 6) form a triangle &nbs
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